A secant line intersects the graph of $h(x)=\sin(x)$ at two points with $x$ -coordinates $4$ and $t$. What is the slope of the secant line? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sin\left(t-4\right)}{t-4}$ (Choice B) B $\dfrac{\sin(t)-\sin\left(4\right)}{4}$ (Choice C) C $\dfrac{\sin\left(t-4\right)}{t}$ (Choice D) D $\dfrac{\sin(t)-\sin\left(4\right)}{t-4}$
Solution: We are given that the secant line intersects the graph at $x=4$ and $x=t$. Since these points are on the graph of $h(x)=\sin(x)$, we know that they must be $\left( 4, \sin(4)\right)$ and $(t, \sin(t))$, respectively. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{\sin(t)-\sin(4)}{t-4} \end{aligned}$ In conclusion, the slope of the secant line is $\dfrac{\sin(t)-\sin(4)}{t-4}$.